HA8-P4

[Victor Ivrii]

Problem 4
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.P.html#problem-6.P.4

Use Fourier method rather than formula

[Emily Deibert]

As in problem three, we have a general solution (before imposing the boundary condition) given by http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.4.html#mjx-eqn-eq-6.4.10. Now we must differentiate this general solution in order to make use of the boundary conditions. This yields: \begin{equation}

u_r(r. \theta{}) = \sum_{n=1}^{\infty}nr^{n-1}(A_n\cos(n\theta{}) + C_n\sin(n\theta{})) \end{equation}

Evaluating this at a, as per the boundary condition: \begin{equation}

u_r(a, \theta{}) = \sum_{n=1}^{\infty}na^{n-1}(A_n\cos(n\theta{}) + C_n\sin(n\theta{})) = f(\theta) \end{equation}

Using the Fourier method, we find the forms of the coefficients to be: \begin{equation}

A_n = \frac{1}{\pi{}na^{n-1}}\int_0^{2\pi{}}f(\theta{}')\cos(n\theta{}')d\theta{}' = \frac{1}{\pi{}na^{n-1}}\Big[\int_0^{\pi{}}\cos(n\theta{}')d\theta{}' - \int_{\pi}^{2\pi{}}\cos(n\theta{}')d\theta{}'\Big]  = 0 \end{equation}

and

\begin{equation}

C_n = \frac{1}{\pi{}na^{n-1}}\int_0^{2\pi{}}f(\theta{}')\sin(n\theta{}')d\theta{}' = \frac{1}{\pi{}na^{n-1}}\Big[\int_0^{\pi{}}\sin(n\theta{}')d\theta{}' - \int_{\pi}^{2\pi{}}\sin(n\theta{}')d\theta{}'\Big] \longrightarrow \end{equation} \begin{equation} = \frac{1}{\pi{}na^{n-1}}\Big[\frac{1-\cos(n\pi)}{n}-\frac{\cos(n\pi)-\cos(2n\pi)}{n}\Big]= \frac{1}{\pi{}na^{n-1}} \frac{2(1-\cos(n\pi{}))}{n}\end{equation}

So: \begin{equation}
C_n = \frac{2(1-\cos(n\pi{}))}{\pi{}n^{2}a^{n-1}}
\end{equation}

As in the previous problem, only terms with odd n survive. This results in: \begin{equation}

C_n = \frac{4}{\pi{}n^{2}a^{n-1}} \end{equation}

Now, plugging these coefficients back into the general solution, we arrive at the solution: \begin{equation}

u(r, \theta{}) = \frac{4}{\pi{}} \sum_{n \geq 1, odd} \frac{r^n\sin(n\theta{})}{n^{2}a^{n-1}} \end{equation}

[Emily Deibert]

(b) Part b is similar, but we consider equations outside the disk. This time the general solution is given by  http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.4.html#mjx-eqn-eq-6.4.11. Much of the solution is the same as part (a) so I will omit several steps. Note that the derivative in this case is: \begin{equation}

u_r(r. \theta{}) = \sum_{n=1}^{\infty}-nr^{-n-1}(B_n\cos(n\theta{}) + D_n\sin(n\theta{})) \end{equation}

As with the $A_n$ terms in part (a), the $B_n$ terms will go to zero. The $D_n$ terms will become: \begin{equation}

D_n = \frac{-a^{n+1}2(1-\cos(n\pi{}))}{\pi{}n^{2}} \end{equation}

Only odd-n terms survive; so: \begin{equation}

D_n = \frac{-4a^{n+1}}{\pi{}n^2} \end{equation}

Plugging into the general solution, we get a final answer: \begin{equation}

u(r, \theta{}) = -\frac{4}{\pi{}} \sum_{n \geq 1, odd} \frac{a^{n+1} \sin(n \theta{})}{n^2 r^n}

\end{equation}

[Emily Deibert]

By the way, we can only satisfy the problem if $h(\theta)$ is such that $\int_0^{2\pi}h(\theta)d\theta = 0$. This is clearly satisfied in this case.